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Musya8 [376]
3 years ago
6

What is the modulus of resilience for a tensile test specimen with a nearly linear elastic region if the yield strength is 500MP

a at a strain of 0.003
Engineering
2 answers:
Temka [501]3 years ago
4 0

Answer:

U = 0.75 MPa

Therefore, the modulus of resilience for the tensile test specimen is 0.75MPa

Explanation:

The Modulus of resilience can be derived from the area under the elastic portion of the stress-strain diagram. It can be written mathematically as;

U = 1/2 × yield strength × yield strain.

Where U is the modulus of resilience.

Given;

Yield strength = 500MPa

Yield strain = 0.003

Substituting the given values we have;

U = 1/2 × 500MPa × 0.003

U = 0.75 MPa

Therefore, the modulus of resilience for the tensile test specimen is 0.75MPa

ella [17]3 years ago
3 0

Answer:

The modulus of resilience is 166.67 MPa

Explanation:

Modulus of resilience is given by yield strength ÷ strain

Yield strength = 500 MPa

Strain = 0.003

Modulus of resilience = 500 MPa ÷ 0.003 = 166.67 MPa

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A steam reformer operating at 650C and 1 atm uses propane as fuel for hydrogen production. At the given operating conditions, th
dusya [7]

Answer:

Explanation:

a) for shifting reactions,

Kps =  ph2 pco2/pcoph20

=[h2] [co2]/[co] [h2o]

h2 + co2 + h2O + co + c3H8 = 1

it implies that

H2 + 0.09 + H2O + 0.08 + 0.05 = 1

solving the system of equation yields

H2 = 0.5308,

H2O = 0.2942

B)  according to Le chatelain's principle for a slightly exothermic reaction, an increase in temperature favors the reverse reaction producing less hydrogen. As a result, concentration of hydrogen in the reformation decreases with an increasing temperature.

c) to calculate the maximum hydrogen yield , both reaction must be complete

C3H8 + 3H2O ⇒ 3CO + 7H2( REFORMING)

CO + H2O ⇒ CO2 + H2 ( SHIFTING)

C3H8 + 6H2O ⇒ 3CO2 + 10 H2 ( OVER ALL)

SO,

Maximum hydrogen yield

= 10mol h2/3 molco2 + 10molh2

= 0.77

⇒ 77%

3 0
3 years ago
Technician A says that the carpet padding is designed to help reduce noise and vibrations.
Firdavs [7]

Answer:

Technicians A is right for the answer

4 0
2 years ago
A gas flows through a one-inlet, one-exit control volume operating at steady state. Considering an adiabatic control volume with
Hunter-Best [27]

Answer:

b. equal to the specific entropy of the gas at the inlet.

Explanation:

Isentropic process is the process in which the entropy of the system remains unchanged. The word isentropic is formed from the combination of the prefix "iso" which means "equal" and the word entropy.

If a process is completely reversible, without the need to provide energy in the form of heat, then the process is isentropic.

3 0
3 years ago
What is the resultant force on one side of a 25cm diameter circular plate standing at the bottom of 3m of pool water?
Tom [10]

Answer:

F=1.47 KN

Explanation:

Given that

Diameter of plate = 25 cm

Height of pool h = 3 m

We know that force can be given as

F= P x A

P=ρ x g x h

Now by putting the values

P=1000 x 10 x 3

P= 30 KPa

A=\dfrac{\pi}{4}\times 0.25^2\ m^2

A=0.049\ m^2

F= 30 x 0.049 KN

F=1.47 KN

So the force on the plate will be 1.47 KN.

4 0
2 years ago
A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A
OLEGan [10]

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos \frac{28.655*S}{R} )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos \frac{28.65 *s }{678}  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + \frac{u^2}{30(\frac{a}{3.2} )-G1}  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + \frac{u^2}{30[(11.2/32.2)-0 ]}

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

5 0
2 years ago
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